Classical-Mechanics · Unit 13 · Video 2 · Interactive Practice

Same Throw, Two Answers: Relative Velocity on a Circle and Recoil from a Rolling Cart

IKey Formulas

Formula	Name	What you need
$\vec{r}\,' = \vec{r} - \vec{R} \qquad \vec{v} = \vec{v}\,' + \vec{V}$	Frame relations — law of addition of velocities	$\vec{R}$ , the position of the moving frame
$\vec{v}\,' = 2\omega\ell\cos\theta\,\hat{i}$	Velocity of $a$ relative to $b$ on the circle	$\hat{r}_a,\ \hat{r}_b$ resolved into fixed $\hat{i},\hat{j}$
$v_{f,\mathrm{cart}} = \dfrac{m_3 v_0\cos\theta}{m_1+m_2+m_3}$	Recoil speed from $p_x$ conservation	The throw converted into the ground frame
$\tan\varphi = \dfrac{m_1+m_2+m_3}{m_1+m_2}\,\tan\theta$	Launch angle measured from the ground	Both ground-frame components of the ball

Key Insight: A quantity measured in a moving frame must pass through $\vec{v} = \vec{v}\,' + \vec{V}$ before it enters a conservation law written in the ground frame — at the instant of release the cart is already sliding backward, so the ball's ground-frame $x$ -velocity is $v_0\cos\theta - v_{f,\mathrm{cart}}$ , not $v_0\cos\theta$ .

IITwo Particles on a Circle

Both particles keep the same speed $\omega\ell$ , yet their relative speed swings between $2\omega\ell$ and zero.

θ = ωt 0°

IIIRecoil from the Cart

The ground frame sees the same throw with less run and the same rise — how much less?

Ball mass m₃ 20 kg

💡 Vertical momentum is not conserved here: to launch the ball upward the person pushes down on the cart, so the ground pushes back with $N > (m_1+m_2+m_3)g$ during the throw.

IVThe Angle the Ground Sees

Is there any pair of masses for which the ground observer sees a shallower launch than the thrower aimed?

Ball mass m₃ 20 kg

💡 When $m_3 \gg m_1+m_2$ the light cart is flung backward at nearly $v_0\cos\theta$ , leaving the heavy ball only its vertical velocity $v_0\sin\theta$ — which is why $\varphi \to 90\degree$ instead of the ball simply flying farther.

VQuiz Questions

Problem 1 · Relative Speed on the Circle

Given: particles $a$ and $b$ run in opposite directions around a circle of radius $\ell$ with angular speed $\omega$ , both starting at the top. When each has swept $\theta = 60\degree$ from the vertical — find the speed of $a$ relative to $b$ .

✅ Correct!

|\vec{v}\,'| = 2\omega\ell\cos 60\degree = 2\omega\ell\cdot\tfrac{1}{2} = \omega\ell

— the particles are still separating, but at only half the rate they had at the top.

❌ That is the value at $\theta = 0$ .

2\omega\ell

is the relative speed at the instant of launch, where

\cos\theta = 1

; by

60\degree

the cosine has fallen to

\tfrac{1}{2}

❌ You used the sine.

2\ell\sin\theta

is the separation

|\vec{r}\,'|

; differentiating it brings down a cosine, so the relative velocity carries

\cos\theta

❌ Equal speeds are not equal velocities. The relative velocity vanishes only where the two velocity vectors coincide — at

\theta = 90\degree

, where both point straight down.

Show solution

Step 1 — write both positions with rotating unit vectors. Put frame $S$ at the centre and let $S'$ ride with $b$ , so $\vec{R} = \ell\hat{r}_b$ and $\vec{r} = \ell\hat{r}_a$ :

\vec{r}\,' = \vec{r} - \vec{R} = \ell\hat{r}_a - \ell\hat{r}_b

Step 2 — resolve the rotating unit vectors into fixed components (the move that makes the problem tractable). With $\theta = \omega t$ measured from the vertical and the particles mirrored across it:

\hat{r}_a = \sin\theta\,\hat{i} + \cos\theta\,\hat{j}, \qquad \hat{r}_b = -\sin\theta\,\hat{i} + \cos\theta\,\hat{j}

\vec{r}\,' = \ell\big(2\sin\theta\big)\hat{i} = 2\ell\sin\theta\,\hat{i}

The $\cos\theta\,\hat{j}$ terms cancel: the separation is always horizontal.

Step 3 — differentiate. Only $\theta$ depends on time, and $d\theta/dt = \omega$ :

\vec{v}\,' = \frac{d}{dt}\big(2\ell\sin\theta\big)\hat{i} = 2\ell\cos\theta\,\frac{d\theta}{dt}\,\hat{i} = 2\omega\ell\cos\theta\,\hat{i}

Step 4 — evaluate at $\theta = 60\degree$ .

|\vec{v}\,'| = 2\omega\ell\cos 60\degree = \omega\ell

Each particle's own speed is the constant $\omega\ell$ , yet the relative speed runs from $2\omega\ell$ at the top down to $0$ at the sides: relative velocity depends on directions, not just speeds.

Problem 2 · Which Momentum Equation Is Legal

Given: a person ( $m_1$ ) on a frictionless cart ( $m_2$ ) throws a ball ( $m_3$ ) at speed $v_0$ and angle $\theta$ as measured on the cart; the cart recoils with speed $v_{f,\mathrm{cart}}$ in the $-\hat{i}$ direction. Find the correct statement of $x$ -momentum conservation in the ground frame.

✅ Correct! Every velocity in the equation is now a ground-frame velocity: the ball's is

\vec{v}\,'_{\!f,\mathrm{ball}} + \vec{V}

, and

\vec{V} = -v_{f,\mathrm{cart}}\hat{i}

❌ This is the frame slip the video warns about.

v_0\cos\theta

is measured on the cart, but conservation was written in the ground frame — the cart is already sliding backward when the ball leaves the hand.

❌ Right idea, wrong sign for $\vec{V}$ . The cart moves in the

-\hat{i}

direction, so adding the frame velocity subtracts

v_{f,\mathrm{cart}}

from the ball's horizontal component.

❌ That is the full speed, not its $x$ -component. Only

v_0\cos\theta

belongs in a horizontal momentum balance — and it still has to be converted to the ground frame.

Show solution

Step 1 — the conserved component. Before the throw everything is at rest, so $p_{x,0}^{\mathrm{total}} = 0$ . The wheels are frictionless, so no external horizontal force acts and $p_x$ stays zero.

Step 2 — the cart's contribution. Once the ball is gone the person rides with the cart:

\vec{p}_{f,\mathrm{cart}} = -(m_1+m_2)\,v_{f,\mathrm{cart}}\,\hat{i}

Step 3 — convert the throw. In the cart frame $\vec{v}\,'_{\!f,\mathrm{ball}} = v_0\cos\theta\,\hat{i} + v_0\sin\theta\,\hat{j}$ . The frame itself moves with $\vec{V} = -v_{f,\mathrm{cart}}\hat{i}$ , so

\vec{v}_{f,\mathrm{ball}} = \vec{v}\,'_{\!f,\mathrm{ball}} + \vec{V} = \big(v_0\cos\theta - v_{f,\mathrm{cart}}\big)\hat{i} + v_0\sin\theta\,\hat{j}

The vertical component is untouched — the frame slides horizontally only.

Step 4 — assemble.

0 = -(m_1+m_2)v_{f,\mathrm{cart}} + m_3\big(v_0\cos\theta - v_{f,\mathrm{cart}}\big)

One equation, one unknown. Collecting the recoil terms gives $v_{f,\mathrm{cart}} = \dfrac{m_3 v_0\cos\theta}{m_1+m_2+m_3}$ .

Problem 3 · Put Numbers on the Throw

Given: $m_1 = 60\ \mathrm{kg}$ (person), $m_2 = 20\ \mathrm{kg}$ (cart), $m_3 = 20\ \mathrm{kg}$ (ball), thrown at $v_0 = 10\ \mathrm{m/s}$ and $\theta = 60\degree$ as measured on the cart — find the cart's recoil speed and the ball's horizontal velocity in the ground frame. Use $\cos 60\degree = \tfrac{1}{2}$ .

Recoil speed of the cart

Horizontal velocity of the ball (ground frame)

✅ Correct! Check the books:

20(4.0) = 80

and

(60+20)(1.0) = 80

— the ball carries exactly the horizontal momentum the cart carries backward.

❌ Check the denominator and the cosine. All three masses share the recoil,

m_1+m_2+m_3 = 100\ \mathrm{kg}

, and only the horizontal part

v_0\cos\theta = 5.0\ \mathrm{m/s}

enters. Dropping

m_3

gives

1.25\ \mathrm{m/s}

; dropping

\cos\theta

gives

2.0\ \mathrm{m/s}

❌ Subtract the recoil, don't add it and don't skip it.

5.0\ \mathrm{m/s}

is the cart-frame value,

6.0\ \mathrm{m/s}

flips the sign of

\vec{V}

, and

8.66\ \mathrm{m/s}

is the vertical component

v_0\sin\theta

Show solution

Step 1 — the cart-frame components.

v_0\cos\theta = 10\cdot\tfrac{1}{2} = 5.0\ \mathrm{m/s}, \qquad v_0\sin\theta = 10\cdot\tfrac{\sqrt{3}}{2} = 8.66\ \mathrm{m/s}

Step 2 — recoil speed.

v_{f,\mathrm{cart}} = \frac{m_3 v_0\cos\theta}{m_1+m_2+m_3} = \frac{20 \cdot 5.0}{60+20+20} = \frac{100}{100} = 1.0\ \mathrm{m/s}

Step 3 — the ball in the ground frame.

\big(v_{f,\mathrm{ball}}\big)_x = v_0\cos\theta - v_{f,\mathrm{cart}} = 5.0 - 1.0 = 4.0\ \mathrm{m/s}

Equivalently, using the boxed form, $\dfrac{m_1+m_2}{m_1+m_2+m_3}\,v_0\cos\theta = \dfrac{80}{100}(5.0) = 4.0\ \mathrm{m/s}$ .

Step 4 — verify $p_x = 0$ .

-(80)(1.0) + (20)(4.0) = -80 + 80 = 0\ \checkmark

The vertical component stays at the full $8.66\ \mathrm{m/s}$ , so the ground observer sees the ball leave at $\sqrt{4.0^2 + 8.66^2} = 9.54\ \mathrm{m/s}$ — slower than the $10\ \mathrm{m/s}$ the thrower felt.

Problem 4 · The Angle Disagreement

Given: the same throw ( $m_1+m_2 = 80\ \mathrm{kg}$ , $m_3 = 20\ \mathrm{kg}$ , $\theta = 60\degree$ , $\tan 60\degree = 1.732$ ) — find the launch angle $\varphi$ the ground observer measures, and decide whether any masses could make $\varphi$ smaller than $\theta$ .

The ground-frame launch angle

Could $\varphi < \theta$ for some choice of masses?

✅ Correct! The recoil removes horizontal velocity and leaves the vertical velocity alone, so the ground-frame direction can only tilt up.

❌ Check which way the mass fraction goes. The ball's horizontal component shrinks by the factor

\tfrac{m_1+m_2}{m_1+m_2+m_3}

, so that factor lands in the denominator of

\tan\varphi

and the ratio

\tfrac{100}{80} = 1.25

multiplies

\tan\theta

. Using

0.8\tan\theta

gives

54.2\degree

; using

\tfrac{m_1+m_2+m_3}{m_3} = 5

gives

83.4\degree

❌ Those cases change how big the disagreement is, not its direction. Heavy ball or light ball, the factor multiplying

\tan\theta

1 + \tfrac{m_3}{m_1+m_2}

, which is never less than

1

Show solution

Step 1 — take the ratio of the ground-frame components.

\tan\varphi = \frac{\big(v_{f,\mathrm{ball}}\big)_y}{\big(v_{f,\mathrm{ball}}\big)_x} = \frac{v_0\sin\theta}{\dfrac{m_1+m_2}{m_1+m_2+m_3}\,v_0\cos\theta}

Step 2 — cancel $v_0$ and flip the mass fraction up.

\tan\varphi = \frac{m_1+m_2+m_3}{m_1+m_2}\,\tan\theta

Step 3 — put the numbers in.

\tan\varphi = \frac{100}{80}(1.732) = 1.25(1.732) = 2.165 \quad\Longrightarrow\quad \varphi \approx 65.2\degree

Step 4 — the sign of the disagreement. Write the factor as $1 + \dfrac{m_3}{m_1+m_2}$ . For any positive masses this exceeds $1$ , so $\tan\varphi > \tan\theta$ and $\varphi > \theta$ always — the ground observer always sees a steeper launch than the thrower intended.

Limits. With $m_3 \ll m_1+m_2$ the factor $\to 1$ and $\varphi \to \theta$ : the massive cart barely recoils. With $m_3 \gg m_1+m_2$ , $\tan\varphi$ grows without bound and $\varphi \to 90\degree$ — the heavy ball keeps only $v_0\sin\theta$ while the light cart is flung backward at nearly $v_0\cos\theta$ .

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